Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers MCQs with answers

Answer: (b) CH₃–CH₂–C(CH₃)(OH)–CH₂–CH₃

In CH₃CH₂C(CH₃)(OH)CH₂CH₃ the –OH is on a carbon bonded to three other carbons: tertiary. (a) and (d) have –OH on a –CH₂– carbon (primary); in (c) the –OH carbon carries one hydrogen and two carbons (secondary).

Answer: (d) C₆H₅–CH(OH)–CH₂–CH₃

In a benzylic alcohol the –OH is on an sp³ carbon next to the ring; in (d) the carbinol carbon is joined directly to C₆H₅ (a secondary benzylic alcohol). In (b) the –OH carbon is one carbon away from the ring; (c) is a phenol, –OH on the ring itself; (a) is allylic.

Answer: (c) 2-hydroxy-5-methylbenzaldehyde

This is the Reimer–Tiemann reaction: –CHCl₂ enters ortho to the –O⁻ group and the substituted benzal chloride is hydrolysed by the alkali to –CHO. The para position is already held by CH₃, so –CHO goes ortho to –OH. Numbering from the CHO carbon: CHO at C-1, OH at C-2, and CH₃ (para to OH) at C-5. (b) is the Kolbe product (CO₂, not CHCl₃); (a) loses the –OH; (d) puts CH₃ meta to OH.

Answer: (a) 3-methylpentan-1-ol

The longest chain has five carbons and is numbered from the end nearest –OH: OH on C-1, CH₃ on C-3; the 'e' of pentane becomes 'ol': 3-methylpentan-1-ol. Numbering from the other end (b) gives the –OH a higher locant; (c) uses a four-carbon chain; (d) names –OH as a prefix.

Answer: (b) butane-1,2-diol

For a polyhydric alcohol the 'e' of the alkane is kept and 'di' goes before 'ol'; the chain is numbered to give the –OH groups the lowest locants: butane-1,2-diol. (a) drops the 'e'; (c) numbers from the wrong end; (d) treats –OH as a prefix.

Answer: (c) benzene-1,3-diol

Catechol is benzene-1,2-diol, resorcinol benzene-1,3-diol, and hydroquinone (quinol) benzene-1,4-diol. 3-Methylphenol is m-cresol.

Answer: (d) 1-methoxybutane

In the IUPAC system the larger group (butane) is the parent and –OCH₃ the substituent on C-1: 1-methoxybutane. (b) makes the smaller group the parent; (c) puts the methoxy group on C-2; (a) is an isomer, CH₃CH₂–O–CH₂CH₂CH₃.

Answer: (a) C–O–C in methoxymethane > tetrahedral angle (109°28′) > C–O–H in methanol

In methanol the lone pairs on oxygen squeeze the C–O–H angle to 108.9°, slightly below tetrahedral. In methoxymethane the two bulky alkyl groups repel and open the C–O–C angle to 111.7°, slightly above tetrahedral.

Answer: (d) conjugation of an oxygen lone pair with the ring gives the bond partial double bond character, and the carbon is sp² hybridised

In phenol the –OH is on an sp² ring carbon; the lone pair on oxygen is delocalised into the ring, giving the C–O bond some double bond character. Both effects shorten the bond. Hydrogen bonding is between molecules and does not shorten the C–O bond.

Answer: (b) 2-methylpropan-1-ol

In hydroboration boron adds to the sp² carbon carrying more hydrogens, the =CH₂ end; oxidation replaces boron by –OH: (CH₃)₂CH–CH₂OH, 2-methylpropan-1-ol. This looks like anti-Markovnikov addition of water. (a) is the acid-catalysed hydration product; (c) and (d) have a different carbon skeleton.